How do you graph the parabola #f(x)= -(x+4)^2 + 9# using vertex, intercepts and additional points?

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Answer 1 · 2018-07-16T06:20:43

Vertex #(-4,9)#
y-intercept #(0,-7)#
x-intercepts #(-1,0); (-7,0)#

Given -

#f(x)=-(x+4)^2+9#

#y=-(x+4)^2+9#

Vertex #(-4,9)#
At #x=0; y=-(0+4)^2+9=-16+9=-7#

y-intercept #(0,-7)#

x intercepts
At #y=0; -(x+4)^2+9=0#

#-(x+4)^2=-9#

Multiplying both sides by #-1#

#(x+4)^2=9#

#x+4=+-sqrt9=+-3#

#x=3-4=-1#

#x=-3-4=-7#

x-intercepts #(-1,0); (-7,0)#

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